Problem: There is a railroad in Takahashi Kingdom. The railroad consists of N sections, numbered 1, 2, ..., N, and N+1 stations, numbered 0, 1, ..., N. Section i directly connects the stations i-1 and i. A train takes exactly A_i minutes to run through section i, regardless of direction. Each of the N sections is either single-tracked over the whole length, or double-tracked over the whole length. If B_i = 1, section i is single-tracked; if B_i = 2, section i is double-tracked. Two trains running in opposite directions can cross each other on a double-tracked section, but not on a single-tracked section. Trains can also cross each other at a station.

Snuke is creating the timetable for this railroad. In this timetable, the trains on the railroad run every K minutes, as shown in the following figure. Here, bold lines represent the positions of trains running on the railroad. (See Sample 1 for clarification.)

<image>

When creating such a timetable, find the minimum sum of the amount of time required for a train to depart station 0 and reach station N, and the amount of time required for a train to depart station N and reach station 0. It can be proved that, if there exists a timetable satisfying the conditions in this problem, this minimum sum is always an integer.

Formally, the times at which trains arrive and depart must satisfy the following:

* Each train either departs station 0 and is bound for station N, or departs station N and is bound for station 0.
* Each train takes exactly A_i minutes to run through section i. For example, if a train bound for station N departs station i-1 at time t, the train arrives at station i exactly at time t+A_i.
* Assume that a train bound for station N arrives at a station at time s, and departs the station at time t. Then, the next train bound for station N arrives at the station at time s+K, and departs the station at time t+K. Additionally, the previous train bound for station N arrives at the station at time s-K, and departs the station at time t-K. This must also be true for trains bound for station 0.
* Trains running in opposite directions must not be running on the same single-tracked section (except the stations at both ends) at the same time.

Constraints

* 1 \leq N \leq 100000
* 1 \leq K \leq 10^9
* 1 \leq A_i \leq 10^9
* A_i is an integer.
* B_i is either 1 or 2.

Input

The input is given from Standard Input in the following format:


N K
A_1 B_1
A_2 B_2
:
A_N B_N


Output

Print an integer representing the minimum sum of the amount of time required for a train to depart station 0 and reach station N, and the amount of time required for a train to depart station N and reach station 0. If it is impossible to create a timetable satisfying the conditions, print -1 instead.

Examples

Input

3 10
4 1
3 1
4 1


Output

26


Input

1 10
10 1


Output

-1


Input

6 4
1 1
1 1
1 1
1 1
1 1
1 1


Output

12


Input

20 987654321
129662684 2
162021979 1
458437539 1
319670097 2
202863355 1
112218745 1
348732033 1
323036578 1
382398703 1
55854389 1
283445191 1
151300613 1
693338042 2
191178308 2
386707193 1
204580036 1
335134457 1
122253639 1
824646518 2
902554792 2


Output

14829091348

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
#define fi first
#define se second
#define pii pair<int,int>
#define mp make_pair
#define pb push_back
#define space putchar(' ')
#define enter putchar('\n')
#define eps 1e-10
#define MAXN 100005
//#define ivorysi
using namespace std;
typedef long long int64;
typedef unsigned int u32;
typedef double db;
template<class T>
void read(T &res) {
    res = 0;T f = 1;char c = getchar();
    while(c < '0' || c > '9') {
    	if(c == '-') f = -1;
    	c = getchar();
    }
    while(c >= '0' && c <= '9') {
    	res = res * 10 +c - '0';
    	c = getchar();
    }
    res *= f;
}
template<class T>
void out(T x) {
    if(x < 0) {x = -x;putchar('-');}
    if(x >= 10) {
    	out(x / 10);
    }
    putchar('0' + x % 10);
}

int N;
int64 K,A[MAXN],sum[MAXN];
int B[MAXN],tot;
int64 L[MAXN],R[MAXN],dpL[MAXN],dpR[MAXN],val[MAXN];
struct node {
    int l,r;
    int mx,lz;
}tr[MAXN * 4];
int id;
void update(int u) {
    if(!id) tr[u].mx = min(tr[u << 1].mx,tr[u << 1 | 1].mx);
    else tr[u].mx = max(tr[u << 1].mx,tr[u << 1 | 1].mx);
}
void addlz(int u,int v) {
    tr[u].lz = v;
    tr[u].mx = v;
}
void pushdown(int u) {
    if(tr[u].lz) {
        addlz(u << 1,tr[u].lz);
        addlz(u << 1 | 1,tr[u].lz);
        tr[u].lz = 0;
    }
}
void build(int u,int l,int r,int v) {
    tr[u].l = l;tr[u].r = r;tr[u].lz = 0;
    if(l == r) {
        tr[u].mx = v;
        return;
    }
    int mid = (l + r) >> 1;
    build(u << 1,l,mid,v);
    build(u << 1 | 1,mid + 1,r,v);
    update(u);
}
void cover(int u,int l,int r,int v) {
    if(l > r) return;
    if(tr[u].l == l && tr[u].r == r) {
        addlz(u,v);
        return;
    }
    pushdown(u);
    int mid = (tr[u].l + tr[u].r) >> 1;
    if(r <= mid) cover(u << 1,l,r,v);
    else if(l > mid) cover(u << 1 | 1,l,r,v);
    else {cover(u << 1,l,mid,v);cover(u << 1 | 1,mid + 1,r,v);}
    update(u);
}
int query(int u,int l,int r) {
    if(l > r) return N + 1;
    if(tr[u].l == l && tr[u].r == r) return tr[u].mx;
    pushdown(u);
    int mid = (tr[u].l + tr[u].r) >> 1;
    if(r <= mid) return query(u << 1,l,r);
    else if(l > mid) return query(u << 1 | 1,l,r);
    else {
        if(!id) return min(query(u << 1,l,mid),query(u << 1 | 1,mid + 1,r));
        else return max(query(u << 1,l,mid),query(u << 1 | 1,mid + 1,r));
    }
}
int64 dist(int64 A,int64 B) {
    if(B >= A) return B - A;
    else return K - (A - B);
}
void Solve() {
    read(N);read(K);
    for(int i = 1 ; i <= N ; ++i) {
        read(A[i]);read(B[i]);
        sum[i] = sum[i - 1] + A[i];
        if(B[i] == 1) {
            if(2 * A[i] > K) {puts("-1");return;}
            L[i] = (-2 * sum[i - 1] % K + K) % K;
            R[i] = (-2 * sum[i] % K + K) % K;
            val[++tot] = L[i];val[++tot] = R[i];
        }
    }

    sort(val + 1,val + tot + 1);
    tot = unique(val + 1,val + tot + 1) - val - 1;
    id = 0;
    build(1,1,tot,N + 1);
    for(int i = N ; i >= 1 ; --i) {
        if(B[i] == 1) {
            int l = lower_bound(val + 1,val + tot + 1,L[i]) - val;
            int r = lower_bound(val + 1,val + tot + 1,R[i]) - val;
            int t = query(1,l,l);
            if(t == N + 1) dpL[i] = 0;
            else dpL[i] = min(dpL[t] + dist(L[i],L[t]),dpR[t] + dist(L[i],R[t]));
            t = query(1,r,r);
            if(t == N + 1) dpR[i] = 0;
            else dpR[i] = min(dpR[t] + dist(R[i],R[t]),dpL[t] + dist(R[i],L[t]));
            if(l <= r) {cover(1,1,l - 1,i);cover(1,r + 1,tot,i);}
            else {cover(1,r + 1,l - 1,i);}
        }
    }
    id = 1;
    build(1,1,tot,0);
    int64 ans = 1e18;
    for(int i = 1 ; i <= N ; ++i) {
        if(B[i] == 1) {
            int l = lower_bound(val + 1,val + tot + 1,L[i]) - val;
            int r = lower_bound(val + 1,val + tot + 1,R[i]) - val;
            int t = query(1,l,l);
            if(t == 0) ans = min(ans,dpL[i]);
            t = query(1,r,r);
            if(t == 0) ans = min(ans,dpR[i]);
            if(l <= r) {cover(1,1,l - 1,i);cover(1,r + 1,tot,i);}
            else {cover(1,r + 1,l - 1,i);}
        }
    }
    ans += 2 * sum[N];
    out(ans);enter;
}
int main() {
#ifdef ivorysi
    freopen("f1.in","r",stdin);
#endif
    Solve();
}